a point 22 caliber rifle fires a 1.8 gram bullet at a 400m/s into a block of wood. If the bullet penetrates to a depht of 10cm find the desaceleration of the bullet. The force exerheted on the bullet, and the time.
To find the deceleration of the bullet, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s since the bullet comes to rest)
u = initial velocity (400 m/s)
a = deceleration of the bullet
s = displacement (0.10 m)
Substituting the known values into the equation:
0^2 = (400 m/s)^2 + 2a(0.10 m)
Simplifying:
0 = 160,000 m^2/s^2 + 0.20a
-160,000 m^2/s^2 = 0.20a
Dividing by 0.20:
-800,000 m^2/s^2 = a
So, the deceleration of the bullet is -800,000 m^2/s^2.
To find the force exerted on the bullet, we can use Newton's second law of motion:
F = ma
where:
F = force exerted on the bullet
m = mass of the bullet (1.8 grams = 0.0018 kg)
a = deceleration of the bullet (-800,000 m^2/s^2)
Substituting the known values into the equation:
F = (0.0018 kg)(-800,000 m^2/s^2)
Simplifying:
F = -1,440 N
So, the force exerted on the bullet is -1,440 N.
Finally, to find the time, we can use the equation of motion:
v = u + at
where:
v = final velocity (0 m/s since the bullet comes to rest)
u = initial velocity (400 m/s)
a = deceleration of the bullet (-800,000 m^2/s^2)
t = time
Substituting the known values into the equation:
0 = 400 m/s + (-800,000 m^2/s^2)t
Simplifying:
-400 m/s = -800,000 m^2/s^2 * t
Dividing by -800,000 m^2/s^2:
t = 0.0005 s
So, the time taken for the bullet to decelerate and come to rest is 0.0005 seconds.
To find the deceleration of the bullet, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity of the bullet (0 m/s, since it comes to a stop)
u = initial velocity of the bullet (400 m/s)
a = acceleration (deceleration, in this case)
s = displacement (10 cm = 0.10 m)
Rearranging the equation:
a = (v^2 - u^2) / (2s)
Plugging in the given values:
a = (0^2 - 400^2) / (2 * 0.10)
a = (-160,000) / 0.20
a = -800,000 m/s^2 (negative sign indicates deceleration)
So, the deceleration of the bullet is -800,000 m/s^2.
To calculate the force exerted on the bullet, we can use Newton's second law of motion:
F = m * a
Where:
F = force
m = mass of the bullet (1.8 grams = 0.0018 kg)
a = deceleration (-800,000 m/s^2)
Plugging in the given values:
F = 0.0018 * (-800,000)
F = -1,440 N (negative sign indicates opposing force)
So, the force exerted on the bullet is -1,440 N.
Lastly, to calculate the time it takes for the bullet to come to a stop, we can use the equation:
v = u + at
Where:
v = final velocity (0 m/s)
u = initial velocity (400 m/s)
a = deceleration (-800,000 m/s^2)
t = unknown time
Rearranging the equation:
t = (v - u) / a
Plugging in the given values:
t = (0 - 400) / (-800,000)
t = 0.0005 seconds
So, the time it takes for the bullet to stop is 0.0005 seconds.